# The sum of perimeter of a circle & square is K.where K is some constant.Prove that the sum of their areas is least when the side of square is double the radius of circle.

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by Raman786

i need detail

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by Raman786

i need detail

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Let side of square = a

radius of circle = r

perimeter of square = 4a

perimeter of circle = 2πr

sum = 4a+2πr = K

⇒4a = K-2πr

⇒a = (K-2πr)/4

area of square = a² = ((K-2πr)/4)² = (K-2πr)²/16

area of circle = πr²

sum of areas, S = πr² + (K-2πr)²/16

For minimum area,

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⇒

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⇒

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⇒ (proved)

radius of circle = r

perimeter of square = 4a

perimeter of circle = 2πr

sum = 4a+2πr = K

⇒4a = K-2πr

⇒a = (K-2πr)/4

area of square = a² = ((K-2πr)/4)² = (K-2πr)²/16

area of circle = πr²

sum of areas, S = πr² + (K-2πr)²/16

For minimum area,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ (proved)

p(square)=4s

a.t.q.

2pie r+4a=k

pie r2 +a2

(given that) 2r=a

assuming the value of r=1

then, 2*22/7*1+8(1)=100/7 (sum of peri. )

22/7*(1)square+4(1)square=50/7

so here area is minimum